Let $g \in L^{1}([0, \infty)) \cup C^{1}([0,\infty))$ be a positive and decreasing function. Let $$ \eta(x,s,t) = u(x,t) - u(x,t-s), \ \ (x,s,t) \in (0,L) \times (0,\infty) \times (0,\infty) $$ When I look at $u(x,t)$ as a function of $u(x)$ and $\eta(x,s,t)$ as a function of $\eta(x,s)$ , I have $u(x) \in H_{0}^{1}(0,L)$ and $\eta(x,s) \in H_{0}^{1}(0,L)$. Suppose that $\eta_{s} \in H_{0}^{1}(0,L)$, $\eta(x,0) = 0, \forall x \in (0,L)$ and $$ \int_{0}^{L}\int_{0}^{\infty}g(s)|\partial_{s}\eta_{x}(x,s)|^{2}dsdx = M < \infty $$ Then $$ \int_{0}^{L}\int_{0}^{\infty}g(s)\partial_{s}\eta_{x}(x,s)u_{x}dsdx = -\int_{0}^{L}\int_{0}^{\infty}g^{\prime}(s)\eta_{x}(x,s)u_{x}dsdx $$
My work: For this aim, let $a,b > 0$ (Then I will do $a \to 0$ and $b \to \infty$). Integrating by parts with respect to s, we get $$ \int_{0}^{L}\int_{a}^{b}g(s)\partial_{s}\eta_{x}(x,s)u_{x}dsdx = \int_{0}^{L}\bigg(g(b)\eta_{x}(x,b)u_{x} - g(a)\eta_{x}(x,a)u_{x} - \int_{a}^{b}g^{\prime}(s)\eta_{x}(x,s)u_{x}ds\bigg)dx $$ Then I need to show that $$ \lim_{b \to \infty}\int_{0}^{L}g(b)\eta_{x}(x,b)u(x)dx = 0 \ \ \text{and} \ \ \lim_{a \to 0}\int_{0}^{L}g(a)\eta_{x}(x,a)u(x)dx = 0 $$ Note that as $\eta_{x}(x,0) = 0$ and $g$ is decreasing, then $$ \int_{0}^{L}g(a)\eta_{x}(x,a)u(x)dx = \int_{0}^{L}g(a)\bigg(\int_{0}^{a}\partial_{s}\eta_{x}(x,s)ds \bigg)u(x)dx \leq \int_{0}^{L}\bigg(\int_{0}^{a}g(s)\partial_{s}\eta_{x}(x,s)ds \bigg)u(x)dx $$ Applying Holder's inequality, we get $$ \int_{0}^{L}\bigg(\int_{0}^{a}g(s)\partial_{s}\eta_{x}(x,s)ds \bigg)u(x)dx \leq C a^{\frac{1}{2}} \int_{0}^{L}\int_{0}^{a}g(s)|\partial_{s}\eta_{x}(x,s)|^{2}ds \leq M C a^{\frac{1}{2}} $$ With $C$ depending on the norm $L^{\infty}$ of $g$ and the norm $L^{2}$ of $u$. Then $$ \lim_{a \to 0}\int_{0}^{L}g(a)\eta_{x}(x,a)u(x)dx \leq \lim_{a \to 0} M C a^{\frac{1}{2}} = 0 $$ Unfortunately I am not able to show the other limit.
Like $g(s) \in C^{1}([0,\infty))$ and $\eta_{x}(. ,s) \in L^{2}(0,\infty)$, then $g(s)\eta_{x}(. ,s) \in L^{1}(0, \infty)$. Another thing, like $\eta_{x}(. ,s), \partial_{s}\eta(.,s) \in L^{2}(0,\infty)$, by the Fundamental Theorem of Calculus, $ \eta_{x}(.,s) $ is uniformly continuous, so $\lim_{s \to \infty}\eta_{x}(.,s) =0$ and since $g$ is bounded, it follows that $$ \lim_{b \to \infty}g(b)\eta_{x}(.,b) = 0. $$